[R] Unexpected behaviour of apply()
Hello everyone,
Considering the following code sample :
indexes - function(vec) {
vec - which(vec==TRUE)
return(vec)
}
mat - matrix(FALSE, nrow=10, ncol=10)
mat[1,3] - mat[3,1] - TRUE
Issuing apply(mat, 1, indexes) returns a 10-cell list, as expected.
Now if I do:
mat[1,3] - mat[3,1] - FALSE
apply(mat, 1, indexes)
I would expect a 10-cell list with integer(0) in each cell - instead I get
integer(0), which wrecks my further process.
Is there a simple way to get the result I expect (and the only consistent
one, IMHO) ?
Thanks by advance for your help,
Pierrick Bruneau
http://www.bruneau44.com
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Re: [R] Unexpected behaviour of apply()
Le vendredi 08 mars 2013 à 09:29 +0100, Pierrick Bruneau a écrit :
Hello everyone,
Considering the following code sample :
indexes - function(vec) {
vec - which(vec==TRUE)
return(vec)
}
This is essentially which(), what did you write such a convoluted
function to get the same result?
mat - matrix(FALSE, nrow=10, ncol=10)
mat[1,3] - mat[3,1] - TRUE
Issuing apply(mat, 1, indexes) returns a 10-cell list, as expected.
Now if I do:
mat[1,3] - mat[3,1] - FALSE
apply(mat, 1, indexes)
I would expect a 10-cell list with integer(0) in each cell - instead I get
integer(0), which wrecks my further process.
From ?apply:
If each call to ‘FUN’ returns a vector of length ‘n’, then ‘apply’
returns an array of dimension ‘c(n, dim(X)[MARGIN])’ if ‘n 1’.
If ‘n’ equals ‘1’, ‘apply’ returns a vector if ‘MARGIN’ has length
1 and an array of dimension ‘dim(X)[MARGIN]’ otherwise. If ‘n’ is
‘0’, the result has length 0 but not necessarily the ‘correct’
dimension.
Note especially the last sentence.
Is there a simple way to get the result I expect (and the only consistent
one, IMHO) ?
One of the interests of apply() is that it combines the return values
from all function calls into a convenient form, but this can indeed be a
problem if you cannot know in advance what this form will be. If you
need a list in all cases, then just call lapply():
lapply(seq(nrow(mat)), function(i) which(mat[i,]))
Regards
Thanks by advance for your help,
Pierrick Bruneau
http://www.bruneau44.com
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Re: [R] Unexpected behaviour of apply()
This is nice fodder for 'The R Inferno' -- thanks.
As Milan said, 'which' will suffice as the function.
Here is a specialized function that only returns a
list and is only implemented to work with matrices.
It should solve your current dilemma.
applyL -
function (X, MARGIN, FUN, ...)
{
stopifnot(length(dim(X)) == 2, length(MARGIN) == 1)
FUN - match.fun(FUN)
ans - vector(list, dim(X)[MARGIN])
if(MARGIN == 1) {
for(i in seq_along(ans)) {
ans[[i]] - FUN(X[i,], ...)
}
} else {
for(i in seq_along(ans)) {
ans[[i]] - FUN(X[,i], ...)
}
}
names(ans) - dimnames(X)[[MARGIN]]
ans
}
Pat
On 08/03/2013 08:29, Pierrick Bruneau wrote:
Hello everyone,
Considering the following code sample :
indexes - function(vec) {
vec - which(vec==TRUE)
return(vec)
}
mat - matrix(FALSE, nrow=10, ncol=10)
mat[1,3] - mat[3,1] - TRUE
Issuing apply(mat, 1, indexes) returns a 10-cell list, as expected.
Now if I do:
mat[1,3] - mat[3,1] - FALSE
apply(mat, 1, indexes)
I would expect a 10-cell list with integer(0) in each cell - instead I get
integer(0), which wrecks my further process.
Is there a simple way to get the result I expect (and the only consistent
one, IMHO) ?
Thanks by advance for your help,
Pierrick Bruneau
http://www.bruneau44.com
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Patrick Burns
pbu...@pburns.seanet.com
twitter: @burnsstat @portfolioprobe
http://www.portfolioprobe.com/blog
http://www.burns-stat.com
(home of:
'Impatient R'
'The R Inferno'
'Tao Te Programming')
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Re: [R] Unexpected behaviour of apply()
Thanks for your replies !
The lapply hack does the job in my context, so I'll stick to it (and
actually in any case where I expect variable length results).
About your quote from ?apply : I read it some time ago actually - but with
my recent use in variable length returning FUN, I got fooled...
On Fri, Mar 8, 2013 at 10:43 AM, Patrick Burns pbu...@pburns.seanet.comwrote:
This is nice fodder for 'The R Inferno' -- thanks.
As Milan said, 'which' will suffice as the function.
Here is a specialized function that only returns a
list and is only implemented to work with matrices.
It should solve your current dilemma.
applyL -
function (X, MARGIN, FUN, ...)
{
stopifnot(length(dim(X)) == 2, length(MARGIN) == 1)
FUN - match.fun(FUN)
ans - vector(list, dim(X)[MARGIN])
if(MARGIN == 1) {
for(i in seq_along(ans)) {
ans[[i]] - FUN(X[i,], ...)
}
} else {
for(i in seq_along(ans)) {
ans[[i]] - FUN(X[,i], ...)
}
}
names(ans) - dimnames(X)[[MARGIN]]
ans
}
Pat
On 08/03/2013 08:29, Pierrick Bruneau wrote:
Hello everyone,
Considering the following code sample :
indexes - function(vec) {
vec - which(vec==TRUE)
return(vec)
}
mat - matrix(FALSE, nrow=10, ncol=10)
mat[1,3] - mat[3,1] - TRUE
Issuing apply(mat, 1, indexes) returns a 10-cell list, as expected.
Now if I do:
mat[1,3] - mat[3,1] - FALSE
apply(mat, 1, indexes)
I would expect a 10-cell list with integer(0) in each cell - instead I get
integer(0), which wrecks my further process.
Is there a simple way to get the result I expect (and the only consistent
one, IMHO) ?
Thanks by advance for your help,
Pierrick Bruneau
http://www.bruneau44.com
[[alternative HTML version deleted]]
__**
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/**listinfo/r-helphttps://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/**
posting-guide.html http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Patrick Burns
pbu...@pburns.seanet.com
twitter: @burnsstat @portfolioprobe
http://www.portfolioprobe.com/**blog http://www.portfolioprobe.com/blog
http://www.burns-stat.com
(home of:
'Impatient R'
'The R Inferno'
'Tao Te Programming')
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